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Bayes Theorem: $$Pr(A|B) = \frac{Pr(A)Pr(B|A)}{Pr(B)} \tag{0}$$

Furthermore: $$Pr(A) = Pr(A|B)Pr(B) + Pr(A|\neg B)Pr(|\neg B) \tag{1}$$

Scenario:
Suppose we are not given that an event $B$ has occurred. Assuming we are given a base Probability of $A$: $(Pr(A_1)$ (this is our prior probability), subsequently, we are now given a $Pr(A|B_1)$, a $Pr(B_1)$, a $Pr(A|\neg B_1)$, and a $Pr(\neg B_1)$ as additional information. Assuming, that I now want to calculate a $Pr(A_2)$ that is the posterior probability in light of the new information.

From $(1)$: $$Pr(A_2) = Pr(A|B_1) Pr(B_1) + Pr(A|\neg B_1) Pr(\neg B_1)\tag{2}$$

From $(0)$: $$Pr(A|B_1) = \frac{Pr(A_1)Pr(B_1|A)}{Pr(B_1)} \qquad :\qquad Pr(A|\neg B_1) = \frac{Pr(A_1)Pr(\neg B_1|A)}{Pr(\neg B_1)} \tag{3}$$

Rewriting $(2)$ in light of $(3)$: $$Pr(A_2) = \frac{Pr(A_1)Pr(B_1|A)}{Pr(B_1)} \times Pr(B_1) + \frac{Pr(A_1)Pr(\neg B_1|A)}{Pr(\neg B_1)} \times Pr(\neg B_1)$$

Which gives us: $$Pr(A_2) = Pr(A_1)Pr(B_1|A) + Pr(A_1)Pr(\neg B_1|A)$$ Simplifying to: $$Pr(A_2) = Pr(A_1)\left(Pr(B_1|A) +Pr(\neg B_1|A)\right)\tag{4}$$

Generalising from $(4)$: $$Pr(A_{l+1}) = Pr(A_l)\left(Pr(B_l|A) +Pr(\neg B_l|A)\right)\tag{5}$$

My Question:

  1. Is $(5)$ valid Bayesian inference?
  2. If not, how do I perform valid inference in the specified scenario?
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  • $\begingroup$ I don't understand: has $B_1$ occured or not? The notation $Pr(A_2)$ for the posterior probability is very confusing - the posterior is a probability conditioned to the data you acquired. What data did you acquire? Did $B_1$ occur or not? If you have no data, this is definitely invalid Bayesian inference. $\endgroup$ – DeltaIV May 30 '17 at 11:40
  • $\begingroup$ $B1$ did not occur. We just received new information on $Pr(A|B_1)$ and $Pr(B_1|A)$. $\endgroup$ – Tobi Alafin May 30 '17 at 11:43
  • $\begingroup$ Did I answer your question? $\endgroup$ – DeltaIV Jun 1 '17 at 9:54
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Your equations do not allow one to make any kind of inference (Bayesian or not): they just state that (in your notation) $Pr(A_{l+1})=Pr(A_l)=Pr(A_1)$. Proof:

$$Pr(A_{l+1}) = Pr(A_l)\left(Pr(B_l|A) +Pr(\neg B_l|A)\right)\tag{5}$$

Now, conditional probabilities are probabilities, which means that they follow all the usual probability rules. In particular, since $B_l$ and $\neg B_l$ are mutually exclusive events, we have

$$Pr(B_l|A) +Pr(\neg B_l|A)=Pr(B_l \cup \neg B_l|A)=Pr(\Omega|A)=1$$

Thus $$Pr(A_{l+1}) = Pr(A_l) \cdot 1 = Pr(A_l) = Pr(A_1) $$ Q.E.D.

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